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mixer [17]
3 years ago
11

HELP PLEASE!!!

Mathematics
1 answer:
True [87]3 years ago
6 0

Answer:

The graph of f(x) is shifted k units to the right of the graph of g(x).    

Step-by-step explanation:

Given the function  and  where k <0.

As, horizontal depends on the value of x and are when

g(x)=f(x+h)   graph shifted to left.

g(x)=f(x-h)    graph shifted to right.

Now, given

But here k is negative, when k is considered positive then f(x) becomes

⇒ The graph of f(x) is shifted k units to the right of the graph of g(x).

Correct option is B.

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The diameter of steel rods manufactured on two different extrusion machines is being investigated. Two random samples of sizes
devlian [24]

Answer:

(a) There is no evidence to support the claim that the two machines produce rods with different mean diameters.

P-value is 0.818

(b) 95% confidence interval for the difference in mean rod diameter is (-0.17, 0.27).

This interval shows that the difference in mean is between -0.17 and 0.27.

Step-by-step explanation:

(a) Null hypothesis: The two machines produce rods with the same mean diameter.

Alternate hypothesis: The two machines produce rods with different mean diameter.

Machine 1

mean = 8.73

variance = 0.35

n1 = 15

Machine 2

mean = 8.68

variance = 0.4

n2 = 17

pooled variance = [(15-1)0.35 + (17-1)0.4] ÷ (15+17-2) = 11.3 ÷ 30 = 0.38

Test statistic (t) = (8.73 - 8.68) ÷ sqrt[0.38(1/15 + 1/17)] = 0.05 ÷ 0.218 = 0.23

degree of freedom = n1+n2-2 = 15+17-2 = 30

Significance level = 0.05 = 5%

Critical values corresponding to 30 degrees of freedom and 5% significance level are -2.042 and 2.042.

Conclusion:

Fail to reject the null hypothesis because the test statistic 0.23 falls within the region bounded by the critical values -2.042 and 2.042.

There is no evidence to support the claim that the two machines produce rods with different mean diameters.

Cumulative area of test statistic is 0.5910

The test is a two-tailed test.

P-value = 2(1 - 0.5910) = 2×0.409 = 0.818

(b) Difference in mean = 8.73 - 8.68 = 0.05

pooled sd = sqrt(pooled variance) = sqrt(0.38) = 0.62

Critical value (t) = 2.042

E = t×pooled sd/√n1+n2 = 2.042×0.62/√15+17 = 0.22

Lower limit of difference in mean = 0.05 - 0.22 = -0.17

Upper limit of difference in mean = 0.05 + 0.22 = 0.27

95% confidence interval for the difference in mean rod diameter is between a lower limit of -0.17 and an upper limit of 0.27.

3 0
3 years ago
17. Mrs. Marshall baked 120 cookies. She decided
agasfer [191]

Answer:

20 cookies

Step-by-step explanation:

120*6=20

20 = 1/6 of 120

8 0
3 years ago
Use the formula SA=6s2 , where SA is the surface area and s is the edge length of the cube, to solve this problem.
Irina-Kira [14]
Using the SA formula you'll have to substitute in the edge length for s.

SA = 6(312)²
SA = 584064

The surface area is 584,064 feet²

Hope this helps :)
3 0
2 years ago
Read 2 more answers
Austin checks the temperature at 10:00 p.m. He notices if the temperature
Alecsey [184]

Answer:

A. t - 5 < 26°

Step-by-step explanation:

Record low temperature for that date = 26°

Temperature at 10:00 PM = t

A fall in temperature by 5° = t - 5

Thus, this is said would mean the record low temperature would be breaking at this point, meaning the new temperature low recorded will be less than 26°.

In mathematical expression, this is expressed as;

t - 5 < 26°

4 0
3 years ago
Suppose Kaitlin places $6500 in an account that pays 12% interest compounded each year.
Leya [2.2K]

\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for 1 year}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &1 \end{cases}

\bf A=6500\left(1+\frac{0.12}{1}\right)^{1\cdot 1}\implies A=6500(1.12)\implies A=7280 \\\\[-0.35em] ~\dotfill

\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for 2 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases}

\bf A=6500\left(1+\frac{0.12}{1}\right)^{1\cdot 2}\implies A=6500(1.12)^2\implies A=8153.6

6 0
3 years ago
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